The following notes are from the Ramsey DocCourse in Prague 2016. The notes are taken by me and I have edited them. In the process I may have introduced some errors; email me or comment below and I will happily fix them.

Definitions

In order to analyze what happens to $\text{Aut}(\mathbb{K})$ when $\mathbb{K}$ is not Ramsey, we will introduce the notion of a universal minimal flow, which at its heart is a canonical compact object we can associate to a group. The size (both topologically and in terms of cardinality) of a group’s universal minimal flow will be determined by the “amount of Ramsey” that the group has.

Definition. A $G$-flow $G \curvearrowright X$ in minimal when it has no proper invariant subflow.

Exercise. Show that every $G$-flow admits a minimal subflow. (Use Zorn’s Lemma.)

Definition. $S_\infty$ acts on the set of binary relations on $\mathbb{N}$, i.e. $X = \{0,1\}^{\mathbb{N}^2}$ which is compact. It acts on $X$ by the logic action, where if $R \in X$ then $$\forall x,y \in \mathbb{N}, \forall g \in S_\infty, g \cdot R(x,y) \Leftrightarrow R(g{-1}(x), g{-1}(y)).$$

Exercise. Show that the collection of linear orders $\text{LO}(\mathbb{N})$ is a closed $S_\infty$-invariant subset of $X$ that is minimal. (Where this is with the logic action, and $X$ is defined as above.)

For a fixed $G$, the object that is universal in the class of minimal $G$-flows will be a canonical object we can associate to $G$, called the universal minimal flow of $G$. To make sense of this, we introduce the concept of universality and flow homomorphism.

Definition. Given $G$-flows $G \curvearrowright X$ and $G \curvearrowright Y$, a flow homomorphism is a map $\pi: X \rightarrow Y$ that is continuous and $G$-invariant.

These universal objects always exist, although the proof is non-constructive.

Theorem. Let $G$ be a topological gorup. There is a minimal $G$-flow $G \curvearrowright M(G)$ that is universal in the sense that for all $G \curvearrowright Y$ minimal there is an onto flow homomorphism $\pi: M(G) \rightarrow Y$.

In addition, $M(G)$ is unique (up to flow isomorphism). So $M(G)$ is called the universal minimal flow of $G$.

Typically $M(G)$ will be hard to describe. The following facts show cases where they are easily understood.

Exercise.

$G$ is extremely amenable iff $\vert M(G) \vert = 1$.

If $G$ is compact, then $M(G) = G$. (Here the action is by left translation.)

Two other examples where $M(G)$ is known.

Fact. If $G$ is discrete, then $M(G)$ is any minimal flow of $\beta(G)$.

The first known example of a non-trivial metrizable universal minimal flow is the following.

Theorem (Pestov 1998). If $G= \text{Homeo}(S^1)$, then $M(G) = S^1$, where the action is the natural one.

$M(S_\infty)$

We will compute the universal minimal flow of $S_\infty$. The original proof is due to Glasner-Weiss in 2002, but we will present proof that is easier to generalize. You should compare this with their original proof.

Theorem (Glasner-Weiss, 2002). $M(S_\infty) = \text{LO}(\mathbb{N})$.

Proof. By an earlier exercise, $\text{LO}(\mathbb{N})$ is a minimal flow, so we need “only” show that it is universal. So let $G = S_\infty$ and let $G \curvearrowright X$.

In this way we have that $G^\star = \text{Aut}(\mathbb{N}, <^\mathbb{Q}) \cong \text{Aut}(\mathbb{Q}, <)$ which is extremely amenable by Pestov’s theorem. Note that $G^\star \leq G$. So $G \curvearrowright X$ induces an action $G^\star \curvearrowright X$. By extreme amenability of $G^\star$, there is a $G^\star$-fixed point $x \in X$.

So, in this way we can think of, $\pi: G \cdot <^\mathbb{Q} \rightarrow X$.

Exercise. Show that this map $\pi$ is $G$-equivariant.

Assume for the moment that $\pi$ can be continuously extended to a map $\tilde{\pi}$ on all of $\text{LO}(\mathbb{N})$. In this case $\tilde{\pi}[\text{LO}(\mathbb{N})]$ is a compact subspace of $X$ containing $x$ (the $G^\star$ fixed point), hence $G \cdot x$. Since $X$ is minimal, $X = \overline{G \cdot x} \subseteq \tilde{\pi}[\text{LO}(\mathbb{N})] \subseteq X$. So we are done.

Claim. $\pi$ can be continuously extended to a map $\tilde{\pi}$ on all of $\text{LO}(\mathbb{N})$.

Proof of claim. We would like to show first that $\pi$ is uniformly continuous. What does that even mean in the non-metric setting? How do we capture the interplay between the topology of $\text{LO}(\mathbb{N})$ and the group $G$?

We can’t assume that $X$ has a metric, but it will always have a unique uniformization, which will act like a metric for the purposes of defining uniform continuity.

To extend $\pi$ continuously, if you are familiar with uniform spaces:

Show that $\pi: G \rightarrow X$ is uniformly continuous when $G$ is equipped with $d_R$ (defined in lecture 1).

Show that $\tilde{\pi}: G / G^\star \rightarrow X$ is uniformly continuous when $G/G^\star$ is equipped with the natural projection of $d_R$.

Show that the identification $G/G^\star \cong G \cdot <^\mathbb{Q}$ also holds when $G / G^\star$ is equipped with the natural projection of $d_R$ and $G \cdot <^\mathbb{Q}$ is equipped with the uniform structure from $\text{LO}(\mathbb{N})$.

If you aren’t familiar with uniform spaces, then just pretend that $X$ has a metric and do the same as above.

This part shows why this type of argument doesn’t always work.

Precompact expansions

This proof works directly when you replace $S_\infty$ by $\text{Aut}(\mathbb{K})$ and $(\mathbb{N}, <^\mathbb{Q})$ is replaced by a closed subgroup $G^\star \leq G$ such that

$G^\star$ is extremely amenable, and

$G / G^\star$ is precompact (i.e. the completion is compact) when equipped with the projection of $d_R$.

Question: What does “$G/G^\star$ is precompact” mean combinatorially? Put another way, what do such $G^\star$ look like?

Since $G^\star \leq G = \text{Aut}(\mathbb{K})$ we can think of $G^\star = \text{Aut}(\mathbb{K}^\star)$ as an expansion of $\mathbb{K}$ where $\mathbb{K}^\star = (\mathbb{K}, (R_i^\star)_{i \in I}) = (\mathbb{K}, \vec{R^\star})$, where $I$ is possibly infinite.

If the parity of $R_i^\star$ is denoted by $a(i)$, then $$\vec{R^\star} \in \prod_{i \in I} \{0,1\}^{\mathbb{N}^{a(i)}} =: P^\star$$ is compact.

Here are two exercises to help you understand the interplay of these objects.

A priori, $d^\star$ gives the box topology which could be different than the product topology. However, precompactness guarantees that these are the same.

Exercise. Show that $(G / G^\star, \text{proj}_R)$ is precompact iff $d^\star$ generates the product topology on $P^\star$, and every element of $\text{Age}(\mathbb{K})$ has only finitely many expansions in $\text{Age}(\mathbb{K}^\star)$.

That is, $\mathbb{K}^\star$ is a precompact expansion of $\mathbb{K}$, hence the name.

Recall that $G \curvearrowright X$ is minimal iff there is a flow homomorphism $\pi: X^\star \rightarrow X$. Now for $Y \subseteq X^\star$ any minimal flow we take $y \in Y$ and see that $\pi(y) \supseteq \overline{G \cdot \pi(y)} = X$.

Corollary. Under the same assumptions, any minimal subflow of $\text{Aut}(\mathbb{K}) \curvearrowright X^\star$ is the universal minimal flow.

In particular, $M(\text{Aut}(\mathbb{K}))$ is metrizable.

In practice, computing this requires understanding what the minimal subflows of $\text{Aut}(\mathbb{K}) \curvearrowright X^\star$ look like. This amounts to understanding when $\text{Aut}(\mathbb{K}) \curvearrowright \overline{G \cdot \vec{R^\star}}$ is minimal.